Carcass wrote:

Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability

that neither of the lightbulbs selected will be defective?

Give your answer as a fraction.

Practice Questions

Question: 24

Page: 344

Difficulty: medium

P(neither lightbulb is defective) = P(BOTH lightbulbs work)

= P(1st lightbulb works

AND 2nd lightbulb works)

= P(1st lightbulb works)

x P(2nd lightbulb works)

= 18/20

x 17/19

= 9/10

x 17/19

[since 18/20 = 9/10]= 153/190

Answer: 153/190

ASIDE: P(1st lightbulb works) = 18/20, because there are 20 lightbulbs and 18 of them work.

P(2nd lightbulb works) = 17/19 because,

once we have selected a working bulb for the 1st selection, there are 19 lightbulbs remaining and 17 of them work.

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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